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Home > Operation Research calculators > Simplex method example
2. Simplex method example ( Enter your problem )
( Enter your problem )
  1. Structure of Linear programming problem
  2. Algorithm
  3. Maximization example-1
  4. Maximization example-2
  5. Maximization example-3
  6. BigM method Algorithm
  7. Minimization example-1
  8. Minimization example-2
  9. Minimization example-3
  10. Degeneracy example-1 (Tie for leaving basic variable)
  11. Degeneracy example-2 (Tie first Artificial variable removed)
  12. Unrestricted variable example
  13. Multiple optimal solution example
  14. Infeasible solution example
  15. Unbounded solution example
 		Other related methods
  1. Formulate linear programming model
  2. Graphical method
  3. Simplex method (BigM method)
  4. Two-Phase method
  5. Primal to dual conversion
  6. Dual simplex method
  7. Integer simplex method
  8. Branch and Bound method
  9. 0-1 Integer programming problem
  10. Revised Simplex method
10. Degeneracy example-1 (Tie for leaving basic variable)
(Previous example)		12. Unrestricted variable example
(Next example)

11. Degeneracy example-2 (Tie first Artificial variable removed)


Degeneracy example-2 (Tie - first Artificial variable removed)
During solving LP problem, a situation may arise in which there is a tie between, 2 or more basic variables for leaving the basis. (means minimum ratios are same).
It is called degeneracy and to resolve this we can select any of them arbitrarily.
But if artificial variable is present then it must be removed first.
Example
Find solution using Simplex(BigM) method
MAX z = 750x1 + 900x2 - 450x3
subject to
x1 + 2x2 <= 70
2x1 + 3x2 - x3 <= 100
x1 >= 20
x2 >= 25
and x1,x2,x3 >= 0

Solution:
Problem is
Max `z``=````750``x_1`` + ``900``x_2`` - ``450``x_3`
subject to
`````x_1`` + ``2``x_2`≤`70`
```2``x_1`` + ``3``x_2`` - ````x_3`≤`100`
`````x_1`≥`20`
`````x_2`≥`25`
and `x_1,x_2,x_3 >= 0; `


The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

1. As the constraint-1 is of type '`<=`' we should add slack variable `S_1`

2. As the constraint-2 is of type '`<=`' we should add slack variable `S_2`

3. As the constraint-3 is of type '`>=`' we should subtract surplus variable `S_3` and add artificial variable `A_1`

4. As the constraint-4 is of type '`>=`' we should subtract surplus variable `S_4` and add artificial variable `A_2`

After introducing slack,surplus,artificial variables
Max `z``=````750``x_1`` + ``900``x_2`` - ``450``x_3`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` + ``0``S_4`` - ``M``A_1`` - ``M``A_2`
subject to
`````x_1`` + ``2``x_2`` + ````S_1`=`70`
```2``x_1`` + ``3``x_2`` - ````x_3`` + ````S_2`=`100`
`````x_1`` - ````S_3`` + ````A_1`=`20`
`````x_2`` - ````S_4`` + ````A_2`=`25`
and `x_1,x_2,x_3,S_1,S_2,S_3,S_4,A_1,A_2 >= 0`


Iteration-1 `C_j``750``900``-450``0``0``0``0``-M``-M`
`B``C_B``X_B``x_1` `x_2` Entering variable`x_3``S_1``S_2``S_3``S_4``A_1``A_2`MinRatio
`(X_B)/(x_2)`
`S_1``0``70``1``2``0``1``0``0``0``0``0``(70)/(2)=35`
`S_2``0``100``2``3``-1``0``1``0``0``0``0``(100)/(3)=33.3333`
`A_1``-M``20``1``0``0``0``0``-1``0``1``0`---
 `A_2` Leaving variable`-M``25``0` `(1)`  (pivot element)`0``0``0``0``-1``0``1``(25)/(1)=25``->`
 `z=0` `0=`
`Z_j=sum C_B X_B`		  `Z_j` `Z_j=sum C_B x_j` `-M` `-M=0xx1+0xx2+(-M)xx1+(-M)xx0`
`Z_j=sum C_B x_1`		 `-M` `-M=0xx2+0xx3+(-M)xx0+(-M)xx1`
`Z_j=sum C_B x_2`		 `0` `0=0xx0+0xx(-1)+(-M)xx0+(-M)xx0`
`Z_j=sum C_B x_3`		 `0` `0=0xx1+0xx0+(-M)xx0+(-M)xx0`
`Z_j=sum C_B S_1`		 `0` `0=0xx0+0xx1+(-M)xx0+(-M)xx0`
`Z_j=sum C_B S_2`		 `M` `M=0xx0+0xx0+(-M)xx(-1)+(-M)xx0`
`Z_j=sum C_B S_3`		 `M` `M=0xx0+0xx0+(-M)xx0+(-M)xx(-1)`
`Z_j=sum C_B S_4`		 `-M` `-M=0xx0+0xx0+(-M)xx1+(-M)xx0`
`Z_j=sum C_B A_1`		 `-M` `-M=0xx0+0xx0+(-M)xx0+(-M)xx1`
`Z_j=sum C_B A_2`
`C_j-Z_j` `M+750` `M+750=750-(-M)` `M+900` `M+900=900-(-M)``uarr` `-450` `-450=(-450)-0` `0` `0=0-0` `0` `0=0-0` `-M` `-M=0-M` `-M` `-M=0-M` `0` `0=(-M)-(-M)` `0` `0=(-M)-(-M)`


Positive maximum `C_j-Z_j` is `M+900` and its column index is `2`. So, the entering variable is `x_2`.

Minimum ratio is `25` and its row index is `4`. So, the leaving basis variable is `A_2`.

`:.` The pivot element is `1`.

Entering `=x_2`, Departing `=A_2`, Key Element `=1`

`R_4`(new)`= R_4`(old)

`R_1`(new)`= R_1`(old)`- 2 R_4`(new)

`R_2`(new)`= R_2`(old)`- 3 R_4`(new)

`R_3`(new)`= R_3`(old)

Iteration-2 `C_j``750``900``-450``0``0``0``0``-M`
`B``C_B``X_B` `x_1` Entering variable`x_2``x_3``S_1``S_2``S_3``S_4``A_1`MinRatio
`(X_B)/(x_1)`
`S_1``0` `20` `20=70-2xx25`
`R_1`(new)`= R_1`(old)`- 2 R_4`(new)		 `1` `1=1-2xx0`
`R_1`(new)`= R_1`(old)`- 2 R_4`(new)		 `0` `0=2-2xx1`
`R_1`(new)`= R_1`(old)`- 2 R_4`(new)		 `0` `0=0-2xx0`
`R_1`(new)`= R_1`(old)`- 2 R_4`(new)		 `1` `1=1-2xx0`
`R_1`(new)`= R_1`(old)`- 2 R_4`(new)		 `0` `0=0-2xx0`
`R_1`(new)`= R_1`(old)`- 2 R_4`(new)		 `0` `0=0-2xx0`
`R_1`(new)`= R_1`(old)`- 2 R_4`(new)		 `2` `2=0-2xx(-1)`
`R_1`(new)`= R_1`(old)`- 2 R_4`(new)		 `0` `0=0-2xx0`
`R_1`(new)`= R_1`(old)`- 2 R_4`(new)		`(20)/(1)=20`
 `S_2` Leaving variable`0` `25` `25=100-3xx25`
`R_2`(new)`= R_2`(old)`- 3 R_4`(new)		 `(2)` `2=2-3xx0` (pivot element)
`R_2`(new)`= R_2`(old)`- 3 R_4`(new)		 `0` `0=3-3xx1`
`R_2`(new)`= R_2`(old)`- 3 R_4`(new)		 `-1` `-1=(-1)-3xx0`
`R_2`(new)`= R_2`(old)`- 3 R_4`(new)		 `0` `0=0-3xx0`
`R_2`(new)`= R_2`(old)`- 3 R_4`(new)		 `1` `1=1-3xx0`
`R_2`(new)`= R_2`(old)`- 3 R_4`(new)		 `0` `0=0-3xx0`
`R_2`(new)`= R_2`(old)`- 3 R_4`(new)		 `3` `3=0-3xx(-1)`
`R_2`(new)`= R_2`(old)`- 3 R_4`(new)		 `0` `0=0-3xx0`
`R_2`(new)`= R_2`(old)`- 3 R_4`(new)		`(25)/(2)=12.5``->`
`A_1``-M` `20` `20=20`
`R_3`(new)`= R_3`(old)		 `1` `1=1`
`R_3`(new)`= R_3`(old)		 `0` `0=0`
`R_3`(new)`= R_3`(old)		 `0` `0=0`
`R_3`(new)`= R_3`(old)		 `0` `0=0`
`R_3`(new)`= R_3`(old)		 `0` `0=0`
`R_3`(new)`= R_3`(old)		 `-1` `-1=-1`
`R_3`(new)`= R_3`(old)		 `0` `0=0`
`R_3`(new)`= R_3`(old)		 `1` `1=1`
`R_3`(new)`= R_3`(old)		`(20)/(1)=20`
`x_2``900` `25` `25=25`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `1` `1=1`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `-1` `-1=-1`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		---
 `z=22500` `22500=900xx25`
`Z_j=sum C_B X_B`		  `Z_j` `Z_j=sum C_B x_j` `-M` `-M=0xx1+0xx2+(-M)xx1+900xx0`
`Z_j=sum C_B x_1`		 `900` `900=0xx0+0xx0+(-M)xx0+900xx1`
`Z_j=sum C_B x_2`		 `0` `0=0xx0+0xx(-1)+(-M)xx0+900xx0`
`Z_j=sum C_B x_3`		 `0` `0=0xx1+0xx0+(-M)xx0+900xx0`
`Z_j=sum C_B S_1`		 `0` `0=0xx0+0xx1+(-M)xx0+900xx0`
`Z_j=sum C_B S_2`		 `M` `M=0xx0+0xx0+(-M)xx(-1)+900xx0`
`Z_j=sum C_B S_3`		 `-900` `-900=0xx2+0xx3+(-M)xx0+900xx(-1)`
`Z_j=sum C_B S_4`		 `-M` `-M=0xx0+0xx0+(-M)xx1+900xx0`
`Z_j=sum C_B A_1`
`C_j-Z_j` `M+750` `M+750=750-(-M)``uarr` `0` `0=900-900` `-450` `-450=(-450)-0` `0` `0=0-0` `0` `0=0-0` `-M` `-M=0-M` `900` `900=0-(-900)` `0` `0=(-M)-(-M)`


Positive maximum `C_j-Z_j` is `M+750` and its column index is `1`. So, the entering variable is `x_1`.

Minimum ratio is `12.5` and its row index is `2`. So, the leaving basis variable is `S_2`.

`:.` The pivot element is `2`.

Entering `=x_1`, Departing `=S_2`, Key Element `=2`

`R_2`(new)`= R_2`(old)`-: 2`

`R_1`(new)`= R_1`(old)`- R_2`(new)

`R_3`(new)`= R_3`(old)`- R_2`(new)

`R_4`(new)`= R_4`(old)

Iteration-3 `C_j``750``900``-450``0``0``0``0``-M`
`B``C_B``X_B``x_1``x_2` `x_3` Entering variable`S_1``S_2``S_3``S_4``A_1`MinRatio
`(X_B)/(x_3)`
`S_1``0` `15/2` `15/2=20-25/2`
`R_1`(new)`= R_1`(old)`- R_2`(new)		 `0` `0=1-1`
`R_1`(new)`= R_1`(old)`- R_2`(new)		 `0` `0=0-0`
`R_1`(new)`= R_1`(old)`- R_2`(new)		 `1/2` `1/2=0-(-1/2)`
`R_1`(new)`= R_1`(old)`- R_2`(new)		 `1` `1=1-0`
`R_1`(new)`= R_1`(old)`- R_2`(new)		 `-1/2` `-1/2=0-1/2`
`R_1`(new)`= R_1`(old)`- R_2`(new)		 `0` `0=0-0`
`R_1`(new)`= R_1`(old)`- R_2`(new)		 `1/2` `1/2=2-3/2`
`R_1`(new)`= R_1`(old)`- R_2`(new)		 `0` `0=0-0`
`R_1`(new)`= R_1`(old)`- R_2`(new)		`(15/2)/(1/2)=15`
`x_1``750` `25/2` `25/2=25-:2`
`R_2`(new)`= R_2`(old)`-: 2`		 `1` `1=2-:2`
`R_2`(new)`= R_2`(old)`-: 2`		 `0` `0=0-:2`
`R_2`(new)`= R_2`(old)`-: 2`		 `-1/2` `-1/2=(-1)-:2`
`R_2`(new)`= R_2`(old)`-: 2`		 `0` `0=0-:2`
`R_2`(new)`= R_2`(old)`-: 2`		 `1/2` `1/2=1-:2`
`R_2`(new)`= R_2`(old)`-: 2`		 `0` `0=0-:2`
`R_2`(new)`= R_2`(old)`-: 2`		 `3/2` `3/2=3-:2`
`R_2`(new)`= R_2`(old)`-: 2`		 `0` `0=0-:2`
`R_2`(new)`= R_2`(old)`-: 2`		---
 `A_1` Leaving variable`-M` `15/2` `15/2=20-25/2`
`R_3`(new)`= R_3`(old)`- R_2`(new)		 `0` `0=1-1`
`R_3`(new)`= R_3`(old)`- R_2`(new)		 `0` `0=0-0`
`R_3`(new)`= R_3`(old)`- R_2`(new)		 `(1/2)` `1/2=0-(-1/2)` (pivot element)
`R_3`(new)`= R_3`(old)`- R_2`(new)		 `0` `0=0-0`
`R_3`(new)`= R_3`(old)`- R_2`(new)		 `-1/2` `-1/2=0-1/2`
`R_3`(new)`= R_3`(old)`- R_2`(new)		 `-1` `-1=(-1)-0`
`R_3`(new)`= R_3`(old)`- R_2`(new)		 `-3/2` `-3/2=0-3/2`
`R_3`(new)`= R_3`(old)`- R_2`(new)		 `1` `1=1-0`
`R_3`(new)`= R_3`(old)`- R_2`(new)		`(15/2)/(1/2)=15``->`
`x_2``900` `25` `25=25`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `1` `1=1`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `-1` `-1=-1`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		---
 `z=31875` `31875=750xx25/2+900xx25`
`Z_j=sum C_B X_B`		  `Z_j` `Z_j=sum C_B x_j` `750` `750=0xx0+750xx1+(-M)xx0+900xx0`
`Z_j=sum C_B x_1`		 `900` `900=0xx0+750xx0+(-M)xx0+900xx1`
`Z_j=sum C_B x_2`		 `-(M)/(2)-375` `-(M)/(2)-375=0xx1/2+750xx(-1/2)+(-M)xx1/2+900xx0`
`Z_j=sum C_B x_3`		 `0` `0=0xx1+750xx0+(-M)xx0+900xx0`
`Z_j=sum C_B S_1`		 `(M)/(2)+375` `(M)/(2)+375=0xx(-1/2)+750xx1/2+(-M)xx(-1/2)+900xx0`
`Z_j=sum C_B S_2`		 `M` `M=0xx0+750xx0+(-M)xx(-1)+900xx0`
`Z_j=sum C_B S_3`		 `(3M)/(2)+225` `(3M)/(2)+225=0xx1/2+750xx3/2+(-M)xx(-3/2)+900xx(-1)`
`Z_j=sum C_B S_4`		 `-M` `-M=0xx0+750xx0+(-M)xx1+900xx0`
`Z_j=sum C_B A_1`
`C_j-Z_j` `0` `0=750-750` `0` `0=900-900` `(M)/(2)-75` `(M)/(2)-75=(-450)-(-(M)/(2)-375)``uarr` `0` `0=0-0` `-(M)/(2)-375` `-(M)/(2)-375=0-((M)/(2)+375)` `-M` `-M=0-M` `-(3M)/(2)-225` `-(3M)/(2)-225=0-((3M)/(2)+225)` `0` `0=(-M)-(-M)`


Positive maximum `C_j-Z_j` is `(M)/(2)-75` and its column index is `3`. So, the entering variable is `x_3`.

Minimum ratio is `15` and its row index is `3`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `1/2`.

Entering `=x_3`, Departing `=A_1`, Key Element `=1/2`

`R_3`(new)`= R_3`(old)`xx2`

`R_1`(new)`= R_1`(old)`- 1/2 R_3`(new)

`R_2`(new)`= R_2`(old)`+ 1/2 R_3`(new)

`R_4`(new)`= R_4`(old)

Iteration-4 `C_j``750``900``-450``0``0``0``0`
`B``C_B``X_B``x_1``x_2``x_3``S_1``S_2``S_3``S_4`MinRatio
`S_1``0` `0` `0=15/2-1/2xx15`
`R_1`(new)`= R_1`(old)`- 1/2 R_3`(new)		 `0` `0=0-1/2xx0`
`R_1`(new)`= R_1`(old)`- 1/2 R_3`(new)		 `0` `0=0-1/2xx0`
`R_1`(new)`= R_1`(old)`- 1/2 R_3`(new)		 `0` `0=1/2-1/2xx1`
`R_1`(new)`= R_1`(old)`- 1/2 R_3`(new)		 `1` `1=1-1/2xx0`
`R_1`(new)`= R_1`(old)`- 1/2 R_3`(new)		 `0` `0=(-1/2)-1/2xx(-1)`
`R_1`(new)`= R_1`(old)`- 1/2 R_3`(new)		 `1` `1=0-1/2xx(-2)`
`R_1`(new)`= R_1`(old)`- 1/2 R_3`(new)		 `2` `2=1/2-1/2xx(-3)`
`R_1`(new)`= R_1`(old)`- 1/2 R_3`(new)
`x_1``750` `20` `20=25/2+1/2xx15`
`R_2`(new)`= R_2`(old)`+ 1/2 R_3`(new)		 `1` `1=1+1/2xx0`
`R_2`(new)`= R_2`(old)`+ 1/2 R_3`(new)		 `0` `0=0+1/2xx0`
`R_2`(new)`= R_2`(old)`+ 1/2 R_3`(new)		 `0` `0=(-1/2)+1/2xx1`
`R_2`(new)`= R_2`(old)`+ 1/2 R_3`(new)		 `0` `0=0+1/2xx0`
`R_2`(new)`= R_2`(old)`+ 1/2 R_3`(new)		 `0` `0=1/2+1/2xx(-1)`
`R_2`(new)`= R_2`(old)`+ 1/2 R_3`(new)		 `-1` `-1=0+1/2xx(-2)`
`R_2`(new)`= R_2`(old)`+ 1/2 R_3`(new)		 `0` `0=3/2+1/2xx(-3)`
`R_2`(new)`= R_2`(old)`+ 1/2 R_3`(new)
`x_3``-450` `15` `15=15/2xx2`
`R_3`(new)`= R_3`(old)`xx2`		 `0` `0=0xx2`
`R_3`(new)`= R_3`(old)`xx2`		 `0` `0=0xx2`
`R_3`(new)`= R_3`(old)`xx2`		 `1` `1=1/2xx2`
`R_3`(new)`= R_3`(old)`xx2`		 `0` `0=0xx2`
`R_3`(new)`= R_3`(old)`xx2`		 `-1` `-1=(-1/2)xx2`
`R_3`(new)`= R_3`(old)`xx2`		 `-2` `-2=(-1)xx2`
`R_3`(new)`= R_3`(old)`xx2`		 `-3` `-3=(-3/2)xx2`
`R_3`(new)`= R_3`(old)`xx2`
`x_2``900` `25` `25=25`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `1` `1=1`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `0` `0=0`
`R_4`(new)`= R_4`(old)		 `-1` `-1=-1`
`R_4`(new)`= R_4`(old)
 `z=30750` `30750=750xx20+(-450)xx15+900xx25`
`Z_j=sum C_B X_B`		  `Z_j` `Z_j=sum C_B x_j` `750` `750=0xx0+750xx1+(-450)xx0+900xx0`
`Z_j=sum C_B x_1`		 `900` `900=0xx0+750xx0+(-450)xx0+900xx1`
`Z_j=sum C_B x_2`		 `-450` `-450=0xx0+750xx0+(-450)xx1+900xx0`
`Z_j=sum C_B x_3`		 `0` `0=0xx1+750xx0+(-450)xx0+900xx0`
`Z_j=sum C_B S_1`		 `450` `450=0xx0+750xx0+(-450)xx(-1)+900xx0`
`Z_j=sum C_B S_2`		 `150` `150=0xx1+750xx(-1)+(-450)xx(-2)+900xx0`
`Z_j=sum C_B S_3`		 `450` `450=0xx2+750xx0+(-450)xx(-3)+900xx(-1)`
`Z_j=sum C_B S_4`
`C_j-Z_j` `0` `0=750-750` `0` `0=900-900` `0` `0=(-450)-(-450)` `0` `0=0-0` `-450` `-450=0-450` `-150` `-150=0-150` `-450` `-450=0-450`


Since all `C_j-Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=20,x_2=25,x_3=15`

Max `z = 30750`



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